Consider the parametric curve: $\begin{aligned} x&=\ln(8t) \\\\ y&=5t^{3} \end{aligned}$ Which integral gives the arc length of the curve over the interval from $t=11$ to $t=17$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\int_{11}^{17} \sqrt{\dfrac{1}{t}+15t^2}\,dt$ (Choice B) B $\int_{11}^{17} \sqrt{(\ln(8t))^2+25t^6}\,dt$ (Choice C) C $\int_{11}^{17} \sqrt{\dfrac{1}{t^2}+225t^4}\,dt$ (Choice D) D $\int_{11}^{17} \sqrt{\dfrac{1}{64t^2}+225t^4}\,dt$
Solution: This is the formula for the arc length of a parametric curve over the interval $[a, b]$ : $\int_a^b\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\, dt$ [Where does this formula come from?] Let's find $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ : $\begin{aligned} \dfrac{dx}{dt}&=\dfrac{d}{dt}\left[\ln(8t)\right] \\\\ &=\dfrac{8}{8t} \\\\ &=\dfrac{1}{t} \\\\\\ \dfrac{dy}{dt}&=\dfrac{d}{dt}\left[5t^{3}\right] \\\\ &=15t^2 \end{aligned}$ Now we can find the expression for the arc length: $\begin{aligned} &\phantom{=}\int_{11}^{17} \sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\,dt \\\\ &=\int_{11}^{17} \sqrt{\left(\dfrac{1}{t}\right)^2+\left(15t^2\right)^2}\,dt \\\\ &=\int_{11}^{17} \sqrt{\dfrac{1}{t^2}+225t^4}\,dt \end{aligned}$ In conclusion, this integral gives the arc length of the curve over the interval from $t=11$ to $t=17$ : $\int_{11}^{17} \sqrt{\dfrac{1}{t^2}+225t^4}\,dt$